minmax_element(ExecutionPolicy&&policy, ForwardIt first, ForwardIt last, Compare comp); (4)(C++17 起) 寻找范围[first,last)中最小和最大的元素。 1)用operator<(C++20 前)std::less{}(C++20 起)比较元素。 3)用比较函数comp比较元素。
class MyClass { public: const std::vector<int> v; const std::pair<int, int> minmax_v; MyClass(const std::vector<int> & v_init) : MyClass(v_init, std::minmax_element(v_init.begin(), v_init.end())) {} private: MyClass(const std::vector<int> & v_init, std::pair<std::...
std::minmax_element std::minmax_element Defined in header <algorithm> (1) template< class ForwardIt > std::pair<ForwardIt,ForwardIt> minmax_element( ForwardIt first, ForwardIt last ); (since C++11) (until C++17) template< class ForwardIt > constexpr std::pair<...
std::minmax_element的简单用法 获取一个数组中的最大值和最小值,通过匿名函数声明自定义比较策略。#include <iostream> #include <vector> #include <algorithm> #include <string>#define BUFSIZE 6 using namespace std;typedef struct { std::string name; ...
{ std::vector<double> v = {}; v.push_back(std::nan("")); v.push_back(20.0); v.push_back(std::nan("")); const auto [mn, mx] = std::minmax_element(begin(v), end(v)); return 0; } compile flags: /std:c++20 /EHsc /Z7 ...
以上方法都可以有效地在std::vector中找到最大值。其中,使用std::max_element是最简洁和推荐的方法,因为它利用了标准库的高效实现。如果你需要同时找到最小值和最大值,std::minmax_element是一个不错的选择。手动遍历和范围for循环提供了更多的灵活性,但在性能上可能不如标准库函数。
minmax_element(I first, S last, Comp comp={}, Proj proj={}); (1)(since C++20) template<ranges::forward_rangeR,classProj=std::identity, std::indirect_strict_weak_order< std::projected<ranges::iterator_t<R>, Proj>>Comp=ranges::less> ...
minmax_element template<class ForwardIt> std::pair<ForwardIt, ForwardIt> minmax_element(ForwardIt first, ForwardIt last) { using value_type = typename std::iterator_traits<ForwardIt>::value_type; return std::minmax_element(first, last, std::less<value_type>()); } minmax_element templat...
min_element Return smallest element in range (function template ) max_element Return largest element in range (function template ) minmax_element Return smallest and largest elements in range (function template ) ...
Usingstd::minmax_element()as suggested in the comments above may be faster for large data sets, but may give slightly different results. Thevaluesfor my example above would be the same, but the position of the "max" element would be9since... ...